Question: $\int 3x(6-x^2)\,dx=$ $+C$
Solution: The integrand is the product of two functions: $3x$ and $6-x^2$. Although it is tempting to take the product of their integrals, this would not work. $\int f(x)\cdot g(x)\,dx\neq\int f(x)\,dx \cdot \int g(x)\,dx$ Instead, what we should do is expand the parentheses so we get a nice polynomial. $\int 3x(6-x^2)\,dx=\int (18x-3x^3)\,dx$ Now we can integrate using the reverse power rule, the sum rule, and the constant multiple rule for indefinite integrals. $\begin{aligned} &\phantom{=}\int 3x(6-x^2)\,dx \\\\ &=\int (18x-3x^3)\,dx \\\\ &= 18\int x\,dx -3\int x^{3}\,dx \\\\ &=18\dfrac{x^2}{2} -3\dfrac{x^4}{4}+C \\\\ &=9 x^2 -\dfrac{3}{4} x^4 +C \end{aligned}$ In conclusion, $\int 3x(6-x^2)\,dx=9 x^2 -\dfrac{3}{4} x^4 +C$